Question

4. If the density of seawater at a depth of 10,000 m is 1.071 g/mL and a 25.0 g sample of water from that depth contains 99.7 mg K , what is the molarity of potassium ions in the sample

Answer 1

Answer:

0.11 M.

Explanation:

Hello!

In this case, since the molarity is computed as the moles of solute divided by the volume of solution in L, we first need to compute the moles of potassium in 99.7 mg as shown below:

$n_K=99.7mgK*\frac{1gK}{1000mgK} *\frac{1molK}{39.1gK}=0.00255molK$

Moreover, given the density of the seawater (mixture containing the K and other solutes), we can compute the liters as shown below:

$V=25.0g*\frac{1mL}{1.071g}*\frac{1L}{1000mL}=0.0233L$

Therefore, the molarity turns out:

$M=\frac{0.00255mol}{0.0233L}\\\\M=0.11M$

Best regards!