Answer:
sample B contains the larger density
Explanation:
Given;
volume of sample A, V = 300 mL = 0.3 L
Molarity of sample A, C = 1 M
volume of sample B, V = 145 mL = 0.145 L
Molarity of sample B, C = 1.5 M
molecular mass of sodium chloride, Nacl = 23 + 35.5 = 58.5 g/mol
Molarity is given as;
![C = \frac{moles \ of \ solute, \ mol}{liters \ of \ solvent} \\\\Moles \ of \ solute \ for \ sample \ A = 1 \times 0.3 = 0.3 \ mol\\\\Moles \ of \ solute \ for \ sample \ B = 1.5 \times 0.145 = 0.2175 \ mol](https://tex.z-dn.net/?f=C%20%3D%20%5Cfrac%7Bmoles%20%5C%20of%20%5C%20solute%2C%20%5C%20mol%7D%7Bliters%20%5C%20of%20%5C%20solvent%7D%20%5C%5C%5C%5CMoles%20%5C%20of%20%5C%20solute%20%5C%20for%20%5C%20sample%20%5C%20A%20%3D%201%20%5Ctimes%200.3%20%3D%200.3%20%5C%20mol%5C%5C%5C%5CMoles%20%5C%20of%20%5C%20solute%20%5C%20for%20%5C%20sample%20%5C%20B%20%3D%201.5%20%5Ctimes%200.145%20%3D%200.2175%20%5C%20mol)
The reacting mass for sample A = 0.3mol x 58.5 g/mol = 17.55 g
The reacting mass for sample B = 0.2175 mol x 58.5 g/mol = 12.72 g
The density of sample A ![= \frac{mass}{volume} = \frac{17.55}{0.3} = 58.5 \ g/L](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7Bmass%7D%7Bvolume%7D%20%3D%20%5Cfrac%7B17.55%7D%7B0.3%7D%20%3D%2058.5%20%5C%20g%2FL)
The density of sample B ![= \frac{mass}{volume} = \frac{12.72}{0.145} = 87.72 \ g/L](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7Bmass%7D%7Bvolume%7D%20%3D%20%5Cfrac%7B12.72%7D%7B0.145%7D%20%3D%2087.72%20%5C%20g%2FL)
Therefore, sample B contains the larger density