# Question 7ba14

Jan 5, 2015

The answer is approximately $1041$ $\text{g/L}$.

Since you start with a concentration of 1M, let's assume you have 1 L of solution. That means that you have

M = n/V => n = M * V = ("1 mole")/("L") * 1L = 1 "mole"

This means that you have a mass of $N a C l$ equal to

m_(NaCl) = n * "molar mass" = 1 "mole" * 58.5 "g"/("mol") = 58.5 $\text{g}$

In order to determine its density, you must calculate the percent concentration by mass of the $N a C l$ solution. This is done by dividing the mass of the solute ($N a C l$) by the total mass of the solution, and multiplying by 100%. Water is assumed to have a density of $\text{1 g/mL}$ (1 L of water = 1000 grams)

%"mass" = m_(NaCl)/m_("sol") * 100%= (58.5g)/(58.5 + 1000g) * 100%

%"mass" = 5.53%#

You'd then turn to a density table like this one:

http://www.maelabs.ucsd.edu/mae171/Conc%20vs%20density.pdf

According to this, a $5.53$ percent concentration by mass corresponds to approximately $1041$ $\text{g/L}$. (Notice that the mass of $N a C l$ at that density is 62.478 g, relatively close to 58.5 g, your value).