Respuesta :
Answer: 319 ml
Explanation:
Given : 86 g of [tex]H_3PO_4[/tex] is dissolved in 100 g of solution.
Density of solution = 1.71 g/ml
Volume of solution=[tex]\frac{\text {mass of solution}}{\text {density of solution}}=\frac{100g}{1.71g/ml}=58.5ml[/tex]
moles of [tex]H_3PO_4=\frac{\text {given mass}}{\text {Molar mass}}=\frac{86g}{98g/mol}=0.88mol[/tex]
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution in L}}[/tex] .....(1)
Molarity of standard [tex]H_3PO_4[/tex] solution =[tex] \frac{0.88\times 1000}{58.5ml}=15.04M[/tex]
To calculate the volume of acid, we use the equation given by neutralization reaction:
[tex]M_1V_1=M_2V_2[/tex]
where,
[tex]M_1\text{ and }V_1[/tex] are the molarity and volume of standard acid which is [tex]H_3PO_4[/tex]
[tex]M_2\text{ and }V_2[/tex] are the molarity and volume of diluted acid which is [tex]H_3PO_4[/tex]
We are given:
[tex]M_1=15.04M\\V_1=?mL\\\\M_2=4.00M\\V_2=1200mL[/tex]
Putting values in above equation, we get:
[tex]15.04\times V_1=4.00\times 1200\\\\V_1=319mL[/tex]
Thus 319 ml of the standard [tex]H_3PO_4[/tex] solution are required.
