Saved Bookmarks
| 1. |
The number of moles of solute present in the solutions of I, II and III is respectively I. 500 mL of 0.2 MNaOH II. 200 mL of 0.1N H_2SO_4 III. 6g of urea in 1 kg of water |
|
Answer» 0.1,0.01,0.1 `=(0.2 times 500)/1000=0.1 MOL` II. Normality=n. factor `times` molarity Molarity `(H_2SO_4)=0.1/2=0.05M` Moles of `H_2SO_4=(0.1 times 200)/(1000 times 2)=0.01 M` III. Moles `=(Weight of UREA)/(Mol EC u lar weight of urea )` `=6/60=0.1 m` Hence, option (a) is correct. |
|