A solution containing 500 g of a protein per litre is isotonic with a solution containing 3.42 g sucrose per litre. The molecular mass of protein is 5× 10x , hence x is

Question

A solution containing 500 g of a protein per litre is isotonic with a solution containing 3.42 g sucrose per litre. The molecular mass of protein is 5× 10x , hence x is (A) 2 (B) 3 (C) 4 (D) 5

Tardigrade Admin · Accepted Answer

In case of isotonic solution π 1=π 2 C1=C2 ((w1/m1 v1))p r o t e i n=((w2/m2 v2))s u c r o s e (500/5 × 1 0x × 1)=(3.42/342 × 1) (100/1 0x)=(1/100) 10x=104 x=4

Q. A solution containing 500 g of a protein per litre is isotonic with a solution containing 3.42 g sucrose per litre. The molecular mass of protein is $5 \times 1 0^{x}$ , hence x is

2

3

4

5

In case of isotonic solution
$π_{1} = π_{2}$
$C_{1} = C_{2}$
$(\frac{w _{1}}{m _{1} v _{1}})_{p ro t e in} = (\frac{w _{2}}{m _{2} v _{2}})_{s u crose}$
$\frac{500}{5 \times 1 0 ^{x} \times 1} = \frac{3.42}{342 \times 1}$
$\frac{100}{1 0 ^{x}} = \frac{1}{100}$
$1 0^{x} = 1 0^{4}$
$x = 4$

Q. A solution containing 500 g of a protein per litre is isotonic with a solution containing 3.42 g sucrose per litre. The molecular mass of protein is 5×10x , hence x is

2

3

4

5

In case of isotonic solution π1​=π2​ C1​=C2​ (m1​v1​w1​​)protein​=(m2​v2​w2​​)sucrose​ 5×10x×1500​=342×13.42​ 10x100​=1001​ 10x=104 x=4

Q. A solution containing 500 g of a protein per litre is isotonic with a solution containing 3.42 g sucrose per litre. The molecular mass of protein is $5 \times 1 0^{x}$ , hence x is

In case of isotonic solution
$π_{1} = π_{2}$
$C_{1} = C_{2}$
$(\frac{w _{1}}{m _{1} v _{1}})_{p ro t e in} = (\frac{w _{2}}{m _{2} v _{2}})_{s u crose}$
$\frac{500}{5 \times 1 0 ^{x} \times 1} = \frac{3.42}{342 \times 1}$
$\frac{100}{1 0 ^{x}} = \frac{1}{100}$
$1 0^{x} = 1 0^{4}$
$x = 4$