A solution of 6.54 g of a carbohydrate in 102.5 g of water has a density of 1.024 g/mL and an osmotic pressure of 4.61 atm at 20.0°C. Calculate the molar mass of the carbohydrate.

The molar mass of the carbohydrate : = 304.19 g/mol

Given

6.54 g carbohydrate

102.5 g of water

osmotic pressure of 4.61 atm

T = 20+273=293 K

Required

The molar mass

Solution

General formula:  

π = osmosis pressure (atm)  

M = solution concentration (mol / l)  

R = constant = 0.08205 L atm mol-1 K-1  

T = Temperature (Kelvin)  

Find molarity(M) :

4.61 atm = M . 0.08205 x 293

M = 0.192 mol/L(mol solute per 1 liter solution)

Total mass of solution :

= 6.54 g + 102.5

= 109.04 g

Volume of solution :

= density x mass

= 1.024 g/ml x 109.04 g

= 111.66 ml

= 0.112 L

mol Carbohydrate (solute):

= M x V

= 0.192 x 0.112

= 0.0215 mol

Molar mass of Carbohydrate :

= mass : mol

= 6.54 : 0.0215

= 304.19 g/mol

answer: e.   sour taste

explanation: acids are those substances which dissociates in water to give ions. as they dissociate to give ions, they act as electrolytes and thus are electrically conductive. acids are sour in taste. example: hcl

bases are those substances which dissociates in water to give ions or accepts ions. as they dissociate to give ions, they act as electrolytes and thus are electrically conductive. bases are bitter in taste. bases are slippery in nature.

b. is the answer hoped i