Assuming that seawater is a 3.50 mass % aqueous solution of nacl, what is the molality of seawater?

Hey there!

Consider 100 g of solution:

Mass of NaCl = 3.50% of mass of seawater

( 3.50 / 100 ) * 100 => 3.50 g

Number of moles as shown below:

Molar mass NaCl = 58.44 g/mol

n = Mass / molar mass

n = 3.50 / 58.44 => 0.059 moles of NaCl

Mass of sweater:

Mass of solution -  Mass of NaCl

100 -  3.50 = 96.5 g

96.5 g in Kg :

96.5 / 1000 => 0.0965 Kg

Therefore ,calculate molality  by using  the  following formula:

molality =  number of moles  of solute / mass of solution

molality = 0.059 / 0.0965

molality = 0.61 m

Hope That helps!

the ideal gas law equation is as follows

pv = nrt

where p - pressure

v - volume

n - number of moles

r - universal gas constant

t - temperature

so if the pressure, volume and temperature are already known

we are left with n and r

since r is the universal gas constant that has a known fixed value then r too is known

so we are left with 'n'

once we know temperature volume and pressure

we can find the number of moles of gas present using the ideal gas law equation

this is not my own answer, this is an answer for the same question by the user blahblahmali. i do not claim ownership of this account.

let that person have a turn to speak : )