Question

Working on-board a research vessel somewhere at sea, you have (carefully) isolated 12.5 micrograms (12.5 ×10–6 g) of what you hope is pure saxitoxin (a non-electrolyte) from a poisonous (and quite cross) puffer fish. You dissolve this sample in 3.10 mL of water and determine that the osmotic pressure of the resulting solution is 0.236 torr at 19ºC (760 torr = 1.00 atm). What is the molar mass of the compound?

Answer 1

Answer:

The value is $Z = 311.33 \ g/mol$

Explanation:

From the question we are told that

The mass of saxitoxin is $m = 12.5 mg = 12.5 * 10^{-6} g$

The volume of water is $V = 3.10 mL = 3.10 *10^{-3} L$

The osmotic pressure is $P = 0.236 = \frac{0.236}{760} = 3.105 * 10^{-4} atm$

The temperature is $T = 19^oC = 19 + 273 = 292 \ K$

Generally the osmotic pressure is mathematically represented as

$P = C * T * R$

Here R is the gas constant with value

$R = 0.0821 ( L .atm /mol. K)$

and C is the concentration of saxitoxin

So

$3.105 * 10^{-4} = C * 0.0821 * 292$

$C = 1.295 *10^{-5} mol/L$

Generally the number of moles of saxitoxin is mathematically represented as

$n = C * V$

=> $n = 1.295 *10^{-5} *3.10 *10^{-3}$

=> $n = 4.015 *10^{-8} \ mol$

Generally the molar mass of saxitoxin is mathematically represented as

$Z = \frac{m}{n}$

=> $Z = \frac{12.5 * 10^{-6}}{ 4.015 *10^{-8}}$

=> $Z = 311.33 \ g/mol$