# Dissolving 120gram of urea in 1000 g of water gave a solution of density 1.15g/mL. The molarity of the solution is:A. 1.79MB. 2MC. 2.05MD. 2.22M

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Hint: To solve this question you should have basic knowledge about the term molarity. The molecular formula of urea is $N{{H}_{2}}CON{{H}_{2}}$ it is required for calculating molar mass of urea, also remember this conversion $m=\rho \times V$ as it is used while solving the question.

Molarity is defined as the number of moles of solute dissolved in one litre (or one cubic decimeter) of solution. It is denoted by M.
Molarity, $M=\dfrac{n}{V}$
Here, ‘n’ denotes the number of moles of solute i.e. equal to $\dfrac{mass of solute}{molar mass of solute}$.
‘V’ denotes volume of solution in litre.
Molecular formula of urea is $N{{H}_{2}}CON{{H}_{2}}$.
Molecular mass of urea is =$14+2\times 1+12+16+14+2\times 1=60$
Mass of solute given in question is 120g.
Hence, no of moles of solute is =$\dfrac{mass of solute}{molar mass of solute}=\dfrac{120}{60}=2$
Density of solution as per given in question is 1.15g/mL
Mass of solvent (i.e. mass of water) is 1000gm as given in question.
According to Law of conservation of mass;
${{m}_{solution}}={{m}_{solute}}+{{m}_{solvent}}$
${{\rho }_{solution}}\times {{V}_{solution}}=120+1000$ (using $m=\rho \times V$)
$1.15\times {{V}_{solution}}=1120$
${{V}_{solution}}=\dfrac{1120}{1.15}=974mL=0.974L$
Thus, $M=\dfrac{n}{V}=\dfrac{2}{0.974}=2.05M$
Hence, we can say that the molarity of the solution is 2.05M.

Hence, option C is the correct answer.

Note:
Don’t get confused between molarity and molality. Molarity (M) is defined as the number of moles of solute dissolved in one litre of solution whereas Molality (m) is defined as number of moles of solute per kilogram (kg) of the solvent.
To avoid calculation mistakes, convert all quantities into SI units before calculation.
The molecular formula of urea is $N{{H}_{2}}CON{{H}_{2}}$ and its molecular weight is 60.