What mass of calcium bromide is needed to prepare 150.0 mL of a 3.50 M solution? (Assume that the molecular weight of CaBr2 is 200.618 g/mol)

105g

Use the following formula to calculate the mass of calcium bromide that is needed:

volume of solution × molar concentration × molar mass = mass of solute

The result is 105 g, which has three significant figures, because the term with the least precision (3.50 M) has three significant figures.

Explanation:

a:

by the combination of an unlimited number of elements

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.