Identify the pair of solutions that exhibits | vedclass

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

Question

(A) Osmotic pressure $\pi = CRT$, where $C$ is molarity. For the same temperature $T$, solutions with the same molarity $C$ have the same osmotic pres

Vedclass · Accepted Answer

$6 	ext{ g/L}$ urea solution and $18 	ext{ g/L}$ glucose solution.

Vedclass · Answer

(A) Osmotic pressure $\pi = CRT$, where $C$ is molarity. For the same temperature $T$, solutions with the same molarity $C$ have the same osmotic pressure.Step $1$: Calculate molarity $C = \frac{	ext{mass (g/L)}}{	ext{molar mass (g/mol)}}$.Step $2$: For urea $(M = 60 	ext{ g/mol})$, $C_1 = \frac{6 	ext{ g/L}}{60 	ext{ g/mol}} = 0.1 	ext{ M}$.Step $3$: For glucose $(M = 180 	ext{ g/mol})$, $C_2 = \frac{18 	ext{ g/L}}{180 	ext{ g/mol}} = 0.1 	ext{ M}$.Step $4$: Since $C_1 = C_2$, the osmotic pressures are equal.

Identify the pair of solutions that exhibits the same osmotic pressure. [Molar mass of urea $= 60 \text{ g mol}^{-1}$ and molar mass of glucose $= 180 \text{ g mol}^{-1}$]

Similar Questions

Calculate the mass of a non-volatile solute dissolved in $0.3 \text{ dm}^3$ of water, which has an osmotic pressure of $0.1 \text{ atm}$ at $300 \text{ K}$. [Molar mass of solute = $328 \text{ g mol}^{-1}$, $R = 0.082 \text{ dm}^3 \text{atm K}^{-1} \text{mol}^{-1}$] (in $\text{ g}$)

$A$ $25 \%$ solution of cane-sugar (molar mass $= 342 \ g \ mol^{-1}$) is isotonic with a $5 \%$ solution of a substance $A$. Find the molecular weight of $A$.

Define $Hypertonic$ solution and $Hypotonic$ solution.

At $27^{\circ} C$,the osmotic pressure of a solution containing $4 \ g$ of a non-electrolyte solute in $1.0 \ L$ of solution is $0.4 \ bar$. The molar mass of the solute in $g \ mol^{-1}$ is :
$(R=0.083 \ L \ bar \ K^{-1} \ mol^{-1})$

What is osmotic pressure? Explain and derive its equation.

Vedclass Test Series

Exam Paper Generator

Online Exam Module