If a 5.25% , w/v solution of a substance is i | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) For isotonic solutions,the osmotic pressures are equal: $\pi_1 = \pi_2$.Since the solvent and temperature are the same,the molar concentrations $(

Vedclass · Accepted Answer

$210$

Vedclass · Answer

(D) For isotonic solutions,the osmotic pressures are equal: $\pi_1 = \pi_2$.Since the solvent and temperature are the same,the molar concentrations $(C)$ must be equal: $C_1 = C_2$.The molar concentration is given by $\frac{w \, (g/L)}{Molar \, mass \, (g/mol)}$.For the substance: $C_1 = \frac{5.25 \, g/100 \, mL}{M_1} = \frac{52.5}{M_1} \, mol/L$.For urea: $C_2 = \frac{1.5 \, g/100 \, mL}{60 \, g/mol} = \frac{15}{60} \, mol/L = 0.25 \, mol/L$.Equating the two: $\frac{52.5}{M_1} = 0.25$.$M_1 = \frac{52.5}{0.25} = 210 \, g \, mol^{-1}$.

If a $5.25\% \, w/v$ solution of a substance is isotonic with a $1.5\% \, w/v$ solution of urea $(Molar \, mass = 60 \, g \, mol^{-1})$ in the same solvent,then the molar mass of the substance is ........ $g \, mol^{-1}$. (in $.0$)

Similar Questions

Which of the following solutions has the maximum osmotic pressure?

$A$ $5\%(w/v)$ solution of sucrose is isotonic with a $1\%(w/v)$ solution of substance '$A$'. What is the molar mass of substance '$A$'?

The empirical formula of a non-electrolyte is $CH_2O$. $A$ solution containing $3 \ g$ of the compound exerts the same osmotic pressure as that of $0.05 \ M$ glucose solution. The molecular formula of the compound is

Arrange the following solutions according to decreasing order of osmotic pressure under similar conditions of temperature,assuming complete dissociation:
$A. 0.2 \ m \ KCl$
$B. 0.3 \ m \ MgSO_4$
$C. 0.1 \ m \ BaCl_2$
$D. 0.5 \ m \ Al_2(SO_4)_3$

$A$ $5\% \, w/v$ solution of cane sugar is isotonic with a $1\% \, w/v$ solution of substance $X$. The molecular weight of $X$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module