How many milliliters of 0.303 M NaNO2 contain 1.873 g of NaNO2?
http://www.jiskha.com/display.cgi?id=1429763582
To solve this problem, we need to use the equation:
molarity (M) = moles (mol) / volume (L)
First, we need to find the number of moles of NaNO2 present in 1.873 g. We can use the molecular weight of NaNO2 to convert from grams to moles.
The molecular weight of NaNO2 is:
Na (22.99 g/mol) + N (14.01 g/mol) + 2 * O (16.00 g/mol) = 69.00 g/mol
To find the moles of NaNO2, we divide the mass (in grams) by the molecular weight (in grams/mol):
1.873 g / 69.00 g/mol = 0.02711 mol
Now, we can calculate the volume of the 0.303 M NaNO2 solution using the equation:
0.303 M = 0.02711 mol / V(L)
Rearranging the equation to solve for V:
V(L) = 0.02711 mol / 0.303 M = 0.0894 L = 89.4 mL
Therefore, 89.4 mL of the 0.303 M NaNO2 solution contains 1.873 g of NaNO2.