What weight of anhydrous sodium trioxocarbonate(Iv) is required to prepare exactly 0.05M of its solution in 250cm volumetric flask?

To prepare a 0.05 M solution of sodium trioxocarbonate(IV) in a 250 cm^3 volumetric flask, you need to calculate the amount of anhydrous sodium trioxocarbonate(IV) needed.

First, let's understand what 0.05 M means. It means there are 0.05 moles of the solute (sodium trioxocarbonate(IV)) dissolved per liter of solution.

Given that you want to prepare 250 cm^3 (which is equivalent to 0.25 liters) of the solution, the number of moles required can be calculated using the formula:

$\text{moles}=\text{Molarity}\times \text{Volume (in liters)}$

So, the number of moles needed is:

$\text{moles}=0.05\text{mol/L}\times 0.25\text{L}$ $\text{moles}=0.0125\text{moles}$

Now, we use the molar mass of sodium trioxocarbonate(IV) to convert moles to grams. The molar mass of Na2CO3 is:

$\text{Na2CO3}=2\times \text{Na}+\text{C}+3\times \text{O}$ $=2(22.99)+12.01+3(16.00)$ $=45.98+12.01+48.00$ $=105.99\text{g/mol}$

So, to find the mass needed:

$\text{Mass}=\text{moles}\times \text{Molar mass}$ $\text{Mass}=0.0125\text{moles}\times 105.99\text{g/mol}$ $\text{Mass}\approx 1.324875\text{g}$

Therefore, you need approximately 1.324875 grams of anhydrous sodium trioxocarbonate(IV) to prepare a 0.05 M solution in a 250 cm^3 volumetric flask.