Question

# The molality of $$15 \% (wt./vol.)$$ solution of $$H_{2}SO_{4}$$ of density $$1.1 g/cm^{3}$$ is approximately.

A
1.2
B
1.4
C
1.8
D
1.6

Solution

## The correct option is D $$1.6$$The solution have 15% (mass/volume)let mass of solute = 15gvolume of solution = 100 mlwe have, density = $$1.1 g/cm^3$$so, mass of solution = $$density*volume = 1.1\times 100 = 110g$$$$\text {mass of solvent(g)=} 110-15 g$$molecular mass of $$H_2SO_4= 98$$ so, moles of $$H_2SO_4$$ = $$\dfrac{\text {given weight}}{\text {molecular weight}} = \dfrac{15}{98} = 0.153 moles$$ molality  = $$\dfrac{\text {moles of solute}}{\text {mass of solvent(g)}}\times 1000$$so, molality =$$\dfrac{0.153}{95} \times 1000 = 1.6$$Chemistry

Suggest Corrections

0

Similar questions
View More

People also searched for
View More