Respuesta :
The mass of [tex]Na_2CrO_4[/tex] required will be 0.6075 g.
Calculation of mass:
The reaction:
[tex]Na_2CrO_4 + 2AgNO_3 \rightarrow Ag_2CrO_4 + 2NaNO_3[/tex]
Volume of [tex]AgNO_3[/tex] = 75 mL
Concentration of [tex]AgNO_3[/tex] = 0.100 M
Step 1:
Molarity is moles of solute in 1000 mL of the solution.
[tex]M=\frac{n}{V}\times 1000\\\\ n=\frac{M\times V}{1000} \\\\n=\frac{0.100\times 75}{1000}\\\\n= 0.0075[/tex]
Moles of [tex]AgNO_3[/tex] = 0.0075
Step 2:
From the above equation,
2 moles of [tex]AgNO_3[/tex] is precipitated by 1 mole of [tex]Na_2CrO_4[/tex]
So, 0.0075 moles will be precipitated by-
[tex]\frac{1}{2}\times 0.0075[/tex] moles of [tex]Na_2CrO_4[/tex]
= 0.00375 moles of [tex]Na_2CrO_4[/tex]
Step 3:
[tex]mole=\frac{Given \,weight}{Molar\, mass} \\[/tex]
Molar mass of [tex]Na_2CrO_4[/tex] = 162 g/mol
[tex]0.00375=\frac{x}{162} \\x= 0.00375\times 162\\x= 0.6075 g[/tex]
So, Mass of [tex]Na_2CrO_4[/tex] required will be 0.6075 g.
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