100 mL of 1.5% (w/v) solution of urea is found to have an osmotic pressure of 6.0 atm and 100 mL of 3.42% (w/v) solution of cane sugar is found to have an osmotic pressure of 2.4 atm. If the two solution are mixed the osmotic pressure of the resulting solution in atm is
Assume that there is no reaction between urea and cane sugar )

Options

A. 8.4
B. 16.8
C. 4.2
D. 2.1

Answer

C

Explanation

Firstly, we calculate the osmotic pressure of each solution by using the formula `pi` = CRT and then, calculate the total osmotic pressure.
Given,` w_(1) ("urea")` = 1.5 g
`V_(1) ("urea") ` = 100 mL
`M_(1) = 60 ` g/mol
`w_(2)`(cane sugar) = 3.42 g
`V_(2)`(cane sugar ) = 100 mL
`M_(2)` = 342 g/mol
`pi_(1) ("urea") = (w_(1) xx 100)/(M_(1) xx V_(1)) xx R xx T `
`pi_(1) ("urea") = (1.5 xx 1000 )/(60 xx 200) xx 0.0821 xx 298`
`pi_(1) ("urea") = 3.05 ` atm
Now,
`pi_(2)` (cane sugar ) = `(W_(2))/(M_(2) xx V_(2)) xx 1000 xx R xx T `
` = (3.42 xx 1000)/(342 xx 200) xx 1000 xx 0.0821 xx 298 `
Thus, the osmotic pressure of the resulting solution is ` pi = pi_(1) + pi_(2) = 3.05 + 1.22 = 4.27 ` atm
Thus, the nearest value is in accordance with option (C ).