5. how many grams are in 2.67 × 10²² particles of nah₂po₄?

Answer:

4.44 g

Explanation:

Step1: Calculate molar mass of $NaH_2PO_4$

$M(NaH_2PO_4)=22.99 + 2\times1.01+30.97 + 4\times16.00=119.98\ g/mol$

Step2: Use Avogadro's number

$n=\frac{N}{N_A}$, where $N = 2.67\times 10^{22}$ and $N_A=6.022\times 10^{23}\ mol^{-1}$ $n=\frac{2.67\times 10^{22}}{6.022\times 10^{23}\ mol^{-1}}=0.0443\ mol$

Step3: Calculate mass

$m = n\times M$ $m=0.0443\ mol\times119.98\ g/mol\approx4.44\ g$