Calculation the molarity and molality of a solution made by mixing equal volumes of `35%` by mass of `H_(2)SO_(4)` (density=`1.22 g mL^(-1))` and `65%` by `H_(2)SO_(4)` (density `=1.62 g mL^(-1))`.

Options

A. `8.66 M, 13.3 m`
B. `5.77 M, 10.1 m`
C. `9.86 M, 12.2 m`
D. `7.55 M, 11.1 m`

Answer

4

Explanation

Molality `(m)=("Total moles of " H_(2)SO_(4))/("Total litres of solution")`
Molarity `(m)=("Total moles of " H_(2)SO_(4))/("Total kilogram of solvent")`
Let's consider 100 mL of each type of `H_(2)SO_(4)` in the solution. Therefore, the total volume is `200 mL` or `0.2` litre.
Mass of `100 mL` of `35% H_(2)SO_(4) "soln"="density" xx "volume"`
`=(1.22 g mL^(-1))(100 mL)`
`=122 g`
Similarly, mass of `100 mL` of `65% H_(2)SO_(4) "soln"=(1.62 g mL^(-1))(100 mL)=162 g`
Total mass of soln `=122 g+162 g=284 g`
Mass of `H_(2)SO_(4)` in 35% soln= `35/100xx122 g`
`=42.7 g`
Mass of `H_(2)SO_(4)` in 65% soln`=65/100xx162 g`
`=105.3 g`
Total mass of `H_(2)SO_(4)(solute)=42.7 g+ 105.3 g`
`=148 g`
Thus,
Mass of `H_(2)O (solvent)=(Mass)_("soln")-(Mass)_(solute)`
`=(284 g)-(148 g)`
`=136 g`
Total moles of `H_(2)SO_(4) =("Total mass")/("Molar mass")=(148 g)/(98 g mol^(-1))`
`=1.51 mol`
Thus,
Total molarity (M)`=n_(H_(2)SO_(4))/V_(L)=(1.51 mol)/(0.2 L)`
`=7.55 mol L^(-1)`
Total molality (m)=`n_(H_(2)SO_(4))/g_("solv")xx(1000 g)/(kg)`
`=(1.51 mol)/(136 g)xx(1000 g)/(kg)`
`=11.1 mol kg^(-1)`