How many gramms of glucose is there in 0.2 liters of solution. Osmotic pressure near 37 degress (celsius) is 810.6 kPa.
I tried to solve it this way:
P = 810.6×10³ Pa
V = 2×10^(-4) m³
R = 8.31
T = 310.15 К
М(С₆Р₁₂O₆) = 180
P×V = (m/M)×RT =>
m = (M×P×V) / (R×T)
m = (180 * 810.6×10³×2×10^(-4)) / (8.31 × 310.15)
m = 11.32 gramms
Is it right?