Question

How much glucose must be dissolved in one litre of an aqueous solution so that its osmotic pressure is 2.57 atm at300 K. [R = 0.0821 L atm K-' mol-').​

Answer 1

V= 1L

P=2.57 atm

T=300K

R=0.0821 L atm K-¹mol-¹

PV=nRT

2.57×1/0.0821×300=n

n=0.104 moles

mole=weight/mol wt

0.104×90=weight

weight=9.36g

PLS MARK BRANLIEST IF IT HELPS

Answer 2

Answer:

1.87819732 g

Explanation:

Osmotic pressure =>  pi=CRT

Concentrarion => C = m/v

pi=2.57 atm

T= 300k

R= 0.0821L atm / K*mol

glucose formula = C6 H12 06 =180g(Molar Mass)

volume = 1L=1000ml

                        ACCORDING TO QUESTION

pi = CRT

pi=m/v*R*T

2.57=W/180 /1000 *0.0821*300

W =1.87819732 g

HOPE YOU WILL UNDERSTAND MY SOLUTION

THANK YOU :-)