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Calculate the moality of 1 L solution of...

Calculate the moality of 1 L solution of 93%
`2xx10^(-5)` by w/V `[d_(H_(2)SO_(4))=1.84g//c]`

A

3.71

B

8.5

C

12.4

D

1.042

Text Solution

Verified by Experts

The correct Answer is:
D
Given, `[d_(H_(2))so_(4)= 1.84g//c c]`
Molality (m) `=("Number of moles of solute")/("Weight of solvent in kg").`
Weight = 93 g
Molecular weight = 98g
Solute `= ("weight")/("mol")xx (1)/(" weight of solvent")`
`d = (M)/(V) m = d xx V`
`m = 1.84 " g mL"^(-1)= 1840 " g"`
Weight of `H_(2)SO_(4) = (93)/(100) xx 1000 = 930 " g "`
So, weight of solvent = 1840 - 930 g
= 910 g = 0.91 kg
`= (93 " g ")/(98 " g ")xx (1)/(0. 91 " kg ")`
= 1.042 mol / kg
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