A sample of magnisium metal containing some `MgO` as impurity was dissolved in 125 " mL of " 0.1 N `H_2SO_4`. The volume of `H_2` evolved at `27.3^@C` and 1 atm was 120.1 mL. The resulting solution was found to be 0.02 N with respect to `H_2SO_4`. Calculate (i) the weight of sample dissolved and (ii) the percentage by weight of Mg in the sample. Neglect any change in the volume of the solution (atomic weight of `Mg=24.3`).

A sample of magnisium metal containing some `MgO` as impurity was diss

1.	A sample of magnisium metal containing some `MgO` as impurity was dissolved in 125 " mL of " 0.1 N `H_2SO_4`. The volume of `H_2` evolved at `27.3^@C` and 1 atm was 120.1 mL. The resulting solution was found to be 0.02 N with respect to `H_2SO_4`. Calculate (i) the weight of sample dissolved and (ii) the percentage by weight of Mg in the sample. Neglect any change in the volume of the solution (atomic weight of `Mg=24.3`).
Answer» Volume of `H_2` at `STP=(120.1xx273)/(300.3)=109.2mL` `Mg+H_2SO_4toMgSO_4+H_2` `MgO+H_2SO_4toMgSO_4+H_2O` Weight of Mg`=(24.3)/(22400)xx109.2=0.1185g` `" Eq of "Mg=(0.1185)/(12.15)=0.009753` `" Eq of "H_2SO_4(t otal)=(125xx0.1)/(1000)=0.0125` Excess of `H_2SO_4=(125xx0.02)/(1000)=0.0025` Excess of `H_2SO_4=(125xx0.2)/(1000)=0.0025` `H_2SO_4` used`=0.0125-0.0025=0.01` `=g" Eq of "Mg+g" Eq of "MgO` " Eq of "MgO`=` Total `Eq-" Eq of "Mg` `=0.01-0.009753` `0.000247 " Eq of "MgO` `=0.00247xx(40.3)/(2)g of MgO` `=0.005g` Total weight `=` Weight of `Mg+` weight of `MgO` `=0.1185+0.005=0.1235g` `% of Mg=(0.1185)/(0.1235)xx100=95.5%`

Discussion

No Comment Found

Related InterviewSolutions

The purity of `H_2O_2` in a given sample is `85%.` Calculate the weight of impure sample of `H_2O_2` which requires `10mL` of `M//5 KMnO_4` solution in a titration in acidic mediumA. 2 gB. 0.2 gC. 0.17 gD. 0.15 g
An impure sample of sodium oxalate `(Na_(2)C_(2)O_(4)` weighing 0.20 g is dissolved in aqueous solution of `H_(2)SO_94)` and solution is titrated at `70^(@)C`,requiring 45 mL of 0.02 M `KMnO_(4)` solution. The end point is overrun, and back titration in carried out with 10 mL of 0.1 M oxalic acid solution.Find the purity of `Na_(2)C_(2)O_(4)` in sample:A. 75B. 83.75C. 90.25D. None of these
`60ml` of mixture of equal volumes of `Cl_(2)` and an oxide of chloine, i.e., `Cl_(2)O_(n)` was heated and then cooled back to the original temperature. The resulting gas mixture was found to have volume of `75ml`. On treatment with `KOH` solution, the volume contracted to `15ml`. Assume that all measurements are made at the same temperature and pressure. Deduce the value of `n` in `Cl_(2)O_(n)`. The oxide of `Cl_(2)` n heating decomposes quantiatively to `O_(2)` and `Cl_(2)`.
`12. g` of an impure sample of arsenious oxide was dissolved in water containing `7.5 g` of sodium bicarbonate and the resulting solution was diluted to `250 mL`. `25 mL` of this solution was completely oxidised by `22.4 mL` of a solution of iodine. `25 mL` of this iodine solution reacted with same volume of a solution containing `24.8 g` of sodium thiosulphate `(Na_(2)S_(2)O_(3).5H_(2)O)` in one litre. Calculate teh percentage of arsenious oxide in the sample ( Atomic mass of `As=74`)
`K_(2)Cr_(2)O_(7)` is obtained in the following steps: `2FeCrO_(4)+2Na_(2)CO_(3)+OtoFe_(2)O_(3)+2Na_(2)CrO_(4)+2CO_(2)` `2Na_(2)CrO_(4)+H_(2)SO_(4)toNa_(2)Cr_(2)O_(7)+H_(2)O+Na_(2)SO_(4)` `Na_(2)Cr_(2)O_(7)+2KCltoK_(2)Cr_(2)+2NaCl` To get 0.25 " mol of "`K_(2)Cr_(2)O_(7)`, " mol of "`50%` pure `FeCrO_(4)` required.A. 1 molB. 0.50 molC. 0.25 molD. 0.125 mol
How many moles of `O_(2)` will be liberated by one mole of `CrO_(5)` is the following reaction: `CrO_(5) + H_(2) SO_(4) rarr Cr_(2) (SO_(4))_(3) + H_(2)O + O_(2)`A. `(5)/(2)`B. `(5)/(4)`C. `(9)/(2)`D. none of these
A sample containing 0.4775 of `(NH_(4))_(2)C_(2)O_(4)` and inert material was dissolved in water and made strongly alkaline with KOH which converted `NH_(4)^(o+)` to `NH_(3)` The liberated `NH_(3)` was distilled of `H_(2)SO_(4)` was back titrated with 11.3 " mL of " 0.1214 M NaOH. Calculate (a) `% of (NH_(4))_(2)C_(2)O_(4)=124.10` And atomic weight of `N=14.0078`.
2.0 g sample of NaCN is dissolved in 50 " mL of " 0.3 M mild alkaline `KMnO_(4)` and heated strongly to convert all the `CN^(ɵ)` to `OCN^(ɵ)`. The solution after acidification with `H_(2)SO_(4)` requries 500 " mL of " 0.05 M `FeSO_(4)` Calculate the percentage purity of NaCN in the sample.
10 " mL of " `NaHC_(2)O_(4)` is oxidised by 10 " mL of " 0.02 M `MnO_(4)^(ɵ)`. Hence, 10 " mL of " `NaHC_(2)O_(4)` is neutralised byA. 10 " mL of " 0.1 M NaOHB. 10 " mL of " 0.02 M NaOHC. 10 " mL of " 0.1 M `Ca(OH)_(2)`D. 10 " mL of " 0.05 N `Ba(OH)_(2)`
`CN^(ɵ)` is oxidised by a strong oxidising agent to `NO_3^(ɵ)` and `CO_2` or `CO_3^(2-)` depending upon the acidity of the reaction mixture. `HNO_3` a strong oxidising agent is reduced by a moderate reducing agent to NO. Write the balanced equation of `HNO_3` with KCN. `CH^(ɵ)toCO_2+NO_3^(ɵ)` `NO_3^(ɵ)NO` If this reaction is carried out, what safety precautions are required?

Discussion

No Comment Found

Related InterviewSolutions

Reply to Comment