We perform the calculation and determine the concentration of the sodium carbonate standard solution to be 2.0 moles per liter. We can add this value to our equation for 𝑣. We can multiply the 150 milliliters by one liter over 1000 milliliters. Recognize that 1000 milliliters equals one liter. Our volume is currently in milliliters but must be in liters in order to use the equation. We can substitute this value into our equation.
We perform the calculation and determine the molar mass of sodium carbonate to be 106 grams per mole. The atomic mass of sodium is 23, carbon 12, and oxygen 16. We can find the atomic masses on the periodic table.
H2so4 molar mass plus#
To find the molar mass of sodium carbonate, we will need two times the atomic mass of sodium plus the atomic mass of carbon plus three times the atomic mass of oxygen. The molar mass can be calculated by adding the atomic masses of the constituent atoms. We can substitute the mass given in the problem for lowercase 𝑚, but we will need to solve for the molar mass of sodium carbonate. Then, we can rearrange the equation to solve for the concentration. And capital 𝑀 is the molar mass in grams per mole.Īs we need to know the concentration of the standard solution, we can start by setting our two equations equal to one another. 𝑐 is the concentration in moles per liter. 𝑛 equals 𝑐𝑣 and 𝑛 equals lowercase 𝑚 over capital 𝑀, where 𝑛 in both equations represents the amount in moles. We will also need to recognize two key equations. We can use these two pieces of information to find the concentration of the sodium carbonate standard solution. This sentence describes the making of a standard solution or a solution with a known concentration. 31.8 grams of sodium carbonate was dissolved to make a 150-milliliter solution for a titration. Let’s start this problem by looking at the first sentence. The results showed that 44 milliliters of the Na2CO3 solution completely reacted with an unknown concentration of 150 milliliters of H2SO4. 31.8 grams of Na2CO3 was dissolved to make a 150-milliliter solution for a titration.
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