What volume, in liters, of 1.5 M [tex]$CaCl _2$[/tex] solution can be made using [tex]$1200.0 g CaCl _2$[/tex]?
[tex]
\begin{array}{c}
CaCl_2: 110.98 g / mol \\
{[?] L}
\end{array}
[/tex]
Answer (2)
Using 1200.0 g of C a C l 2 , we can calculate that we can produce approximately 7.21 liters of a 1.5 M C a C l 2 solution by first determining the number of moles and then calculating the volume based on the molarity. This calculation ensures we have the right concentration for experiments. It's essential in laboratory settings to achieve the desired solution concentrations accurately.
;
7.21 L
Explanation
Problem Introduction We are given a mass of C a C l 2 and the molarity of a C a C l 2 solution, and we want to find the volume of the solution in liters that can be made using the given mass.
Calculating Moles of Calcium Chloride First, we need to calculate the number of moles of C a C l 2 using the formula: m o l es = m o l a r ma ss ma ss . We are given the mass of C a C l 2 as 1200.0 g and the molar mass as 110.98 g/mol. Therefore, m o l es = 110.98 g / m o l 1200.0 g = 10.812759055685708 m o l .
Calculating Volume of Solution Next, we need to calculate the volume of the solution using the formula: v o l u m e = m o l a r i t y m o l es . We are given the molarity of the C a C l 2 solution as 1.5 M. Therefore, v o l u m e = 1.5 M 10.812759055685708 m o l = 7.208506037123805 L .
Final Answer Therefore, the volume of 1.5 M C a C l 2 solution that can be made using 1200.0 g C a C l 2 is approximately 7.21 L.
Examples
In chemistry, when preparing solutions for experiments, it's crucial to calculate the correct volume of a solution needed to achieve a specific concentration. For instance, if you're conducting an experiment that requires a 1.5 M C a C l 2 solution and you have 1200.0 g of C a C l 2 , you need to determine the volume of the solution you can make. This calculation ensures that you have the right concentration for your experiment, which directly impacts the accuracy and reliability of your results. Knowing how to perform these calculations is fundamental in many scientific and industrial applications, from pharmaceutical research to environmental monitoring.
