x g of MgSO_4 (i=1.8) in 2.5 L of solution | vedclass

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(D) Given:Mass of $MgSO_4$ $(w) = x \ g$van't Hoff factor $(i) = 1.8$Volume of solution $(V) = 2.5 \ L$Osmotic pressure $(\pi) = 2.463 \ atm$Temperatu

Vedclass · Accepted Answer

$16.6$

Vedclass · Answer

(D) Given:Mass of $MgSO_4$ $(w) = x \ g$van't Hoff factor $(i) = 1.8$Volume of solution $(V) = 2.5 \ L$Osmotic pressure $(\pi) = 2.463 \ atm$Temperature $(T) = 27 + 273 = 300 \ K$Molar mass of $MgSO_4 = 24 + 32 + 64 = 120 \ g \ mol^{-1}$Using the formula for osmotic pressure: $\pi = i 	imes C 	imes R 	imes T = i 	imes \frac{w}{M} 	imes \frac{1}{V} 	imes R 	imes T$Substituting the values: $2.463 = \frac{1.8 	imes x 	imes 0.0821 	imes 300}{120 	imes 2.5}$$	herefore x = \frac{2.463 	imes 120 	imes 2.5}{1.8 	imes 0.0821 	imes 300}$$x = \frac{738.9}{44.334} \approx 16.67 \ g$Rounding to the nearest option,$x = 16.6 \ g$.Hence,option $(d)$ is the correct answer.

$x \ g$ of $MgSO_4$ $(i=1.8)$ in $2.5 \ L$ of solution has an osmotic pressure of $2.463 \ atm$ at $27^{\circ} C$. What is the value of $x$ in $g$?

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