Question

2.00 L of 0.800 M NaNO3 must be prepared from a solution known to be 2.50 M in concentration.How many mL are required? Plus don't send me virus links!!!! pls explain how to do it​

Answer 1

1.36 × 10³ mL of water.

Step-by-step explanation:

We can utilize the dilution equation. Recall that:

`\displaystyle M_1V_1= M_2V_2`

Where M represents molarity and V represents volume.

Let the initial concentration and unknown volume be M₁ and V₁, respectively. Let the final concentration and required volume be M₂ and V₂, respectively. Solve for V₁:

`\displaystyle \begin{aligned} (2.50\text{ M})V_1 &= (0.800\text{ M})(2.00\text{ L}) \\ \\ V_1 & = 0.640\text{ L} \end{aligned}`

Therefore, we can begin with 0.640 L of the 2.50 M solution and add enough distilled water to dilute the solution to 2.00 L. The required amount of water is thus:

`\displaystyle 2.00\text{ L} - 0.640\text{ L} = 1.36\text{ L}`

Convert this value to mL:

`\displaystyle 1.36\text{ L} \cdot \frac{1000\text{ mL}}{1\text{ L}} = 1.36* 10^3\text{ mL}`

Therefore, about 1.36 × 10³ mL of water need to be added to the 2.50 M solution.