Human blood has an osmotic pressure of 7 .65 atm. at body temperature (37°C). Hence molarity of an to have same osmotic pressure,will be

Question

Human blood has an osmotic pressure of 7 .65 atm. at body temperature (37°C). Hence molarity of an to have same osmotic pressure,will be (A) 0.30 M (B) 0.10 M (C) 0.50 M (D) 0.20 M

Tardigrade Admin · Accepted Answer

π V = nRT π = (n/V) RT 7.65 = ( (n/V) ) 0.0821 × 310 (n/V) = (7.65/ 0.081 × 310) = 0.30 mol L-1

Q. Human blood has an osmotic pressure of 7 .65 atm. at body temperature (37 $^{\circ}$ C). Hence molarity of an to have same osmotic pressure,will be

0.30 M

0.10 M

0.50 M

0.20 M

$π V = n RT$
$π = \frac{n}{V} RT$
$7.65 = (\frac{n}{V}) 0.0821 \times 310$
$\frac{n}{V} = \frac{7.65}{0.081 \times 310} = 0.30 m o l L^{- 1}$

Q. Human blood has an osmotic pressure of 7 .65 atm. at body temperature (37∘C). Hence molarity of an to have same osmotic pressure,will be