a chemist must dilute 77.9 ml of 3.08 m aqueous sodium chloride (nacl) solution until the concentration…
a chemist must dilute 77.9 ml of 3.08 m aqueous sodium chloride (nacl) solution until the concentration falls to 2.00 m. hell do this by adding distilled water to the solution until it reaches a certain final volume. calculate this final volume, in liters. round your answer to 3 significant digits.
Answer
Explanation:
Step1: Use dilution formula
The dilution formula is $M_1V_1 = M_2V_2$, where $M_1$ is the initial molarity, $V_1$ is the initial volume, $M_2$ is the final molarity, and $V_2$ is the final volume. Given $M_1 = 3.08\ M$, $V_1=77.9\ mL = 0.0779\ L$, $M_2 = 2.00\ M$. We need to solve for $V_2$. From $M_1V_1 = M_2V_2$, we can express $V_2$ as $V_2=\frac{M_1V_1}{M_2}$.
Step2: Substitute values
Substitute $M_1 = 3.08\ M$, $V_1 = 0.0779\ L$, and $M_2 = 2.00\ M$ into the formula $V_2=\frac{M_1V_1}{M_2}$. $V_2=\frac{3.08\ M\times0.0779\ L}{2.00\ M}$. First, calculate the numerator: $3.08\times0.0779 = 0.239932$. Then, divide by the denominator: $V_2=\frac{0.239932}{2.00}=0.119966\ L$.
Step3: Round to 3 significant digits
Rounding $0.119966\ L$ to 3 significant digits gives $0.120\ L$.
Answer:
$0.120$