Question

05 question (1 point) a 1.00 ml sample of seawater is determined to contain 2.50 × 10-14 moles of dissolved gold. how many grams of gold are in 1.00 l of seawater? 3rd attempt see periodic table see hint

Explanation:

Step1: Convert volume units

Since \(1L = 1000mL\), for a \(1.00L\) sample (which is \(1000mL\)), and we know that in \(1.00mL\) of seawater, the number of moles of gold \(n = 2.50\times10^{-14}\text{mol}\). For \(V = 1000mL\), the number of moles of gold in \(1.00L\) is \(n=2.50\times 10^{-14}\text{mol/mL}\times1000mL=2.50\times 10^{-11}\text{mol}\)

Step2: Use the formula \(m = nM\)

The molar mass of gold \(M = 196.97g/mol\). Using the formula \(m=nM\), where \(n = 2.50\times 10^{-11}\text{mol}\) and \(M=196.97g/mol\). Then \(m=(2.50\times 10^{-11}\text{mol})\times196.97g/mol\approx4.92\times 10^{-9}g\)

Answer:

\(4.92\times 10^{-9}\)