Determine the amount of `CaCl_(2)` (i = 2.47) dissolved in 2.5 L of water sucjh that its osmotic pressure is 0.75 atm at `27^(@)C`.

Options

A. 1.0 g
B. 9.2 g
C. 3.42 g
D. 2.42 g

Answer

c

Explanation

`pi = iCRT pi = "inRT"//V`
`n = (piV)/(iRT) = (0.75atmxx2.5L)/(2.47xx0.0821 L atm K^(-1) mol^(-1)xx300 K)`
`= 0.0308 mol`
Molar mass of `CaCl_(2) = 111 g mol^(-1)`
Amount dissolved `= 0.0308xx111 = 3.42g`