Answer:
1)Molar mass of calcium citrate is 498.46g/mol
4) Mass of [tex]8.94\times 10^{21}[/tex] molecules of ethanol is [tex]68.29\times 10^{-3}g[/tex]
5)926 gm of [tex]FeBr_{3}[/tex] has [tex]18.85\times 10^{23}formula\,units[/tex]
Explanation:
1)
The molecular formula of Calcium citrate- [tex]Ca_{3}(C_{6}H_{5}O_{7})_{2}[/tex]
Let's calculate the molar mass.
[tex]3\times 40g/mol+2((6\times 12)+(5\times1)+(7\times 16))= 498.46g/mol[/tex]
Therefore, molar mass of calcium citrate is 498.46g/mol
4)
One mole of molecule has [tex]6.02\times 10^{23}[/tex] molecules.
Molar mass of ethanol = 46.07 g/mol
[tex]Mass=(8.94\times 10^{21})\times (\frac{1mole}{6.023\times 10^{23}})\times(\frac{46g}{1mole})=68.29\times 10^{-3}g[/tex]
Therefore, mass of [tex]8.94\times 10^{21}[/tex] molecules of ethanol is [tex]68.29\times 10^{-3}g[/tex]
5)
Molar mass of [tex]FeBr_{3}[/tex] = 295.56g/mol
One mole has [tex]6.02\times 10^{23}[/tex] formula units.
Given mass = 926 g
[tex]926 \times \frac{6.023\times 10^{23}\,formula\,units}{295.56}=18.85\times 10^{23}formula\,units[/tex]
Therefore, 926 gm of [tex]FeBr_{3}[/tex] has [tex]18.85\times 10^{23}formula\,units[/tex]
