$M = molarity$ of the solution, $m = molality$ of the solution
$d = density$ of the solution $(in g . {mL}^{- 1}), M^{1} = gram$ molecular weight of solute

Which of the following relations is correct

see full answer

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

m = \frac{M \times 100}{d + {MM}^{1}}

m = \frac{M \times 100}{(1000 \times d) - {MM}^{1}}

m = \frac{M}{1000 d - {MM}^{1}}

M = \frac{m \times 100}{(1000 \times d) - {MM}^{1}}

answer is C.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Weight of solute = n \times M^{1} (g)

$Total weight of solution = 1000 + nM$

$Volume of solution = \frac{1000 + {nM}^{1}}{d}$

$Molarity (M) = \frac{n}{V_{(ml)}} \times 1000$

$= \frac{n}{\frac{100 + {nM}^{1}}{d}} \times 1000$

$M = \frac{nd \times 1000}{1000 + {nM}^{1}}$

$1000 \times M + {nMM}^{1} = nd \times 1000$

$n = \frac{1000 M}{1000 d - {MM}^{1}}$

$m = \frac{n}{kg} = \frac{1000 \times M}{(1000 d - {MM}^{1})}$

Watch 3-min video & get full concept clarity