Solution.
"V_1=100mL=0.1L;"
"c_1=0.500mol\/L;"
"c_2=0.250mol\/L;"
"V_2-?;"
"c_1V_1=c_2V_2,"
"c_1,V_1" - the molarity and volume of the concentrated solution;
"c_2,V_2" - the molarity and volume of the diluted solution;
"V_2=\\dfrac{c_1V_1}{c_2};"
"V_2=\\dfrac{0.500mol\/L\\sdot0.1L}{0.250mol\/L}=0.2L;"
It is impossible to prepare such a solution, because it is necessary to take twice as much acid as the volume of the whole solution should be, because pure acid has twice less concentration than dilute.
Answer: It is not possible to prepare the solution.
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