Question

Calculate the concentration of a solution of glucose which is isotonic at the same temperature with a solution of urea containing 6.2g/L ?​

Answer 1

Answer : The concentration of glucose is, 18.6 g/L

Solution :

Isotonic solutions have same osmotic pressure at same temperature.

$\pi_1=\pi_2\\\\\frac{n_1RT}{V_1}=\frac{n_2RT}{V_2}\\\\\frac{n_1}{V_1}=\frac{n_2}{V_2}\\\\C_1=C_2\\\\\frac{w_1}{M_1V_1}=\frac{w_2}{M_2V_2}$

where,

n = number of moles

w = mass

M = molar mass

V = volume

R = gas constant

T = temperature

Formula used :

$\frac{w_1}{M_1V_1}=\frac{w_2}{M_2V_2}$

where,

$w_1$ = mass of glucose = ?

$V_1$ = volume of glucose = 1 L

$M_1$ = molecular weight of glucose  = 180 g/mole

$w_2$ = mass of urea = 6.2 g

$V_2$ = volume of urea = 1 L

$M_2$ = molecular weight of urea = 60 g/mole

Now put all the given values in the above formula, we get the mass of glucose.

$\frac{w_1}{180g/mole\times 1L}=\frac{6.2g}{60g/mole\times 1L}$

$w_1=18.6g$

That means, 18.6 grams of glucose present in one liter of solution.

Therefore, the concentration of glucose is, 18.6 g/L

Answer 2

Explanation:

answer -18.54 g/l thank you