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# A 0.3%(w/v) solution of urea (mol.wt.60) would be isotonic with

$\begin{array}{1 1}(a)\;0.05M\;glucose\\(b)\;0.1M\;KCl\\(c)\;0.1M\;NaCl\\(d)\;0.3\%\;glucose\; solution\end{array}$

Can you answer this question?

Answer: $0.05\;M\; \text{glucose}$
Molarity (M) is defined as number of moles of solute dissolved in one litre (or one cubic decimetre) of solution, $M = \large\frac{\text{ Moles of solute }}{\text{Molarity Volume of solution in litre}}$
$0.3 \%\; w/v \rightarrow 0.3 \; gm$ of Urea in $100\;mL$ of Water.
Mole of Solute $= \large\frac{0.3}{60} $$= 0.005\; mol Molarity volume of solution = 100\; mL = 0.1\;L \Rightarrow M = \large\frac{0.005}{0.1}$$ = 0.05\;M$
This is identical to the molarity of $0.05\;M\; \text{glucose}$, among the choices given.
answered Feb 13, 2014
edited Jul 16, 2014