Answer : 42.3 ml of a 0.266 M [tex]RbNO_3[/tex] solution are required.
Solution : Given,
Molarity of [tex]RbNO_3[/tex] solution 1 = 0.266 M
Molarity of [tex]RbNO_3[/tex] solution 2 = 0.075 M
Volume of [tex]RbNO_3[/tex] solution 2 = 150 ml = 0.150 L (1 L = 1000 ml)
Formula used :
[tex]M_1V_1=M_2V_2[/tex]
where,
[tex]M_1[/tex] = Molarity of [tex]RbNO_3[/tex] solution 1
[tex]M_2[/tex] = Molarity of [tex]RbNO_3[/tex] solution 2
[tex]V_1[/tex] = Volume of [tex]RbNO_3[/tex] solution 1
[tex]V_2[/tex] = Volume of [tex]RbNO_3[/tex] solution 2
Now put all the given values in above formula, we get
[tex]0.266M\times V_1=0.075M\times 0.150L\\V_1=0.042293L=42.293ml\approx 42.3 ml[/tex] (1 L = 1000 ml)
Therefore, 42.3 ml of a 0.266 M [tex]RbNO_3[/tex] solution are required.