3moles of sodium hydroxide will neutralize 1mole of citric acid,what volume of 2moles of sodium hydroxide solution will neutralize 50cm of citric acid solution?

I don't think you have all of the information needed to solve the problem. You need molarity of the citric acid and molarity of the NaOH. For whatever it's worth, 50 cm of citric acid is not a volume.

To find the volume of the 2 moles of sodium hydroxide solution that will neutralize 50 cm³ of citric acid solution, we can use a simple stoichiometric calculation.

First, let's establish the molar ratio between sodium hydroxide (NaOH) and citric acid (C6H8O7). According to the given information, 3 moles of NaOH will neutralize 1 mole of citric acid.

Next, we need to convert the volume of citric acid solution from cm³ to liters, as moles are usually expressed in liters. Since 1 cm³ is equal to 1 milliliter (ml), we have 50 cm³ = 50 ml. Therefore, the volume of citric acid solution is 50 ml ÷ 1000 ml/L = 0.05 L.

To determine the number of moles of citric acid present in the 0.05 L solution, we need to know its concentration. Let's assume the concentration of the citric acid solution is C mol/L.

Using the molar ratio from earlier, we can set up the following equation:

3 moles NaOH = 1 mole citric acid
2 moles NaOH = x moles citric acid

Since the number of moles of citric acid is directly proportional to its volume, we can set up the following equation:

1 mole citric acid ÷ 0.05 L = x moles citric acid ÷ y L (y is the volume of the 2 M NaOH solution that we need to find)

Simplifying the equation, we get:

1 ÷ 0.05 = x ÷ y

Now, we know that the concentration of the 2 M NaOH solution is 2 mol/L. Therefore, the number of moles of NaOH in the solution can be calculated by multiplying its concentration by the volume (y) in liters:

2 moles/L * y L = 2y moles NaOH

According to the stoichiometric ratio, this amount of NaOH will neutralize the same number of moles of citric acid:

2y moles NaOH = x moles citric acid

Combining the equations:

2y = x ÷ 0.05

Now, we know that 3 moles of NaOH neutralize 1 mole of citric acid. Therefore, x (number of moles of citric acid) will be equal to 0.05 L * (3 moles NaOH/1 mole citric acid) = 0.15 L.

Substituting this value into the equation:

2y = 0.15 ÷ 0.05
2y = 3
y = 3/2
y = 1.5 L

So, the volume of the 2 mol/L NaOH solution required to neutralize 50 cm³ of citric acid solution is 1.5 liters.