Calculate the molarity of each of the following solution : (a) 30g of `CO(NH_(3))_(2).6H_(2)O` in 4.3 L of solution. (b) 30 mL of `0.5" MH"_(2)"SO"_(4)` diluted to 500 mL. (a) Molarity `=("moles of solute")/("Volume of solution litre")` and moles of solute `=("mass of solute")/("molar solution of solute")` So, first find molar mass by adding atomic masses of different elements, then find moles of solute and then molarity. (b) Use molarity equation for dilution. `M_(1)V_(1)=M_(2)V_(2)` (Before dilution) (After dilution)

Calculate the molarity of each of the following solution : (a) 30g of

1.	Calculate the molarity of each of the following solution : (a) 30g of `CO(NH_(3))_(2).6H_(2)O` in 4.3 L of solution. (b) 30 mL of `0.5" MH"_(2)"SO"_(4)` diluted to 500 mL. (a) Molarity `=("moles of solute")/("Volume of solution litre")` and moles of solute `=("mass of solute")/("molar solution of solute")` So, first find molar mass by adding atomic masses of different elements, then find moles of solute and then molarity. (b) Use molarity equation for dilution. `M_(1)V_(1)=M_(2)V_(2)` (Before dilution) (After dilution)
Answer» Molar mass of `CO(NO_(3))_(2).6H_(2)O=(58.7)+2(14+48)+(6xx18)"g mol"^(-1)` `=58.7+124+108=290.7~~291" g mol"^(-1)` Mole of `CO(NO_(3))_(2)6" H"_(2)O=(30"g")/(291" g mol"^(-1))=0.103` Volume of solution = 4.3 L Molarity (M) `=(0.103" mol")/(4.3" L")=0.024" mol L"^(-1)=0.024" M"` (b) Volume of undiluted `H_(2)SO_(4)` solution `(V_(1))=30" mL"` Molarity of undiluted `H_(2)SO_(4)` solution `(M_(1))=0.5" M"` Volume of diluted `H_(2)SO_(4)` solution `(V_(2))=500" mL"` We know that `M_(1)V_(1)=M_(2)V_(2)` `:. M_(2)=(M_(1)V_(1))/(V_(2))=((0.5"M")(30" mL"))/(500" mL")=0.03" M"`

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