A solution is made by dissolving 20 g of a s | vedclass

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

Question

(A) The formula for osmotic pressure is $\pi V = nRT = \frac{W_B}{M_B} RT$.Rearranging for molecular weight $M_B = \frac{W_B 	imes R 	imes T}{\pi

Vedclass · Accepted Answer

$1198$

Vedclass · Answer

(A) The formula for osmotic pressure is $\pi V = nRT = \frac{W_B}{M_B} RT$.Rearranging for molecular weight $M_B = \frac{W_B 	imes R 	imes T}{\pi 	imes V}$.Given: $W_B = 20 \ g$,$V = 500 \ mL = 0.5 \ L$,$T = 15 + 273 = 288 \ K$,$\pi = \frac{600}{760} \ atm$,$R = 0.0821 \ L \ atm \ K^{-1} \ mol^{-1}$.Substituting the values: $M_B = \frac{20 	imes 0.0821 	imes 288}{(600/760) 	imes 0.5}$.$M_B = \frac{20 	imes 0.0821 	imes 288 	imes 760}{600 	imes 0.5} \approx 1198 \ g \ mol^{-1}$.

$A$ solution is made by dissolving $20 \ g$ of a substance in $500 \ mL$ of water. Its osmotic pressure was found to be $600 \ mm$ of $Hg$ at $15 \ ^oC$. The molecular weight of the substance is

Similar Questions

How can you remove the hard calcium carbonate layer of the egg without damaging its semipermeable membrane? Can this egg be inserted into a bottle with a narrow neck without distorting its shape? Explain the process involved.

At $27\,^oC$,the osmotic pressure of a solution containing $36\,g$ of glucose per liter is $4.92\,atm$. If the osmotic pressure of the solution is adjusted to $1.5\,atm$ at the same temperature,what will be its concentration?

$A$ $0.6\%$ solution of urea (molecular weight $= 60$) would be isotonic with

The ideal method to determine osmotic pressure is ...... .

$A$ $5\% \, w/v$ 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