10 g of cane sugar molecular mass - 342 in 1- 10-3 m3 of solution produces an osmotic pressure of 6-68- 104 Nm-2 at 273 K- Calculate the value of R in SI units-

The correct option is A 8.3684 J K^ 1 mol^ 1 Given data: #160; V = 0.001 m^3 = 1 L, π= 6.68×10^4Nm^ 2, T= 273K, w= 10g, M.W.= 342gmol^ 1 We know tha ...

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Byju's Answer

Standard XII

Chemistry

Osmotic Pressure

10 g of cane ...

Question

10 g of cane sugar (molecular mass = 342) in $1 \times 10^{- 3} m^{3}$ of solution produces an osmotic pressure of $6.68 \times 10^{4} N m^{- 2}$ at 273 K. Calculate the value of R in SI units.

8.3684 J K^{- 1} m o l^{- 1}

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9.3684 J K^{- 1} m o l^{- 1}

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7.3684 J K^{- 1} m o l^{- 1}

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5.3684 J K^{- 1} m o l^{- 1}

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Solution

The correct option is A $8.3684 J K^{- 1} m o l^{- 1}$
Given data:
$V = 0.001 m^{3} = 1 L, π = 6.68 \times 10^{4} N m^{- 2}, T = 273 K, w = 10 g, M . W . = 342 g m o l^{- 1}$

We know that Osmotic pressure is-
$π = i C R T$
since, i=1 for cane sugar
$∴ π = C R T$
$6.68 \times 10^{4} = \frac{10}{342} R \times 273$
We get-
$R = 8.36 J K^{- 1} m o l^{- 1}$

Hence, option (A) is correct.

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10 g of cane sugar (molecular mass = 342) in 1×10−3m3 of solution produces an osmotic pressure of 6.68×104Nm−2 at 273 K. Calculate the value of R in SI units.

The correct option is A 8.3684JK−1mol−1Given data: V=0.001 m3=1L,π=6.68×104Nm−2,T=273K,w=10g,M.W.=342gmol−1We know that Osmotic pressure is- π=iCRTsince, i=1 for cane sugar∴π=CRT6.68×104=10342R×273We get-R=8.36JK−1mol−1Hence, option (A) is correct.

10 g of cane sugar (molecular mass = 342) in $1 \times 10^{- 3} m^{3}$ of solution produces an osmotic pressure of $6.68 \times 10^{4} N m^{- 2}$ at 273 K. Calculate the value of R in SI units.