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How many grams of NaOH are needed to prepare 250 mL of 0.205 M NaOH?

Answer 1

Henry Antrim

A bit over #2*g#...............

#"Concentration"# #=# #"Moles of solute (moles)"/"Volume of solution (L)"#

Thus #"Concentration"xx"volume"="moles of solute"#

#0.205*mol*cancel(L^-1)xx0.250*cancelL="Moles of solute"#

#=0.0513*mol#, an answer in moles as required.

And thus #"mass of NaOH"# #=# #"moles "xx" molar mass"#

#=0.0513*cancel(mol)xx40.0*g*cancel(mol^-1)#

#=2.05*g#

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Answer 2

Wyatt Atkins

To calculate the grams of NaOH needed to prepare 250 mL of 0.205 M NaOH solution, you can use the formula:

( \text{grams of NaOH} = \text{volume (L)} \times \text{molarity (mol/L)} \times \text{molar mass (g/mol)} )

First, convert the volume from milliliters to liters:

( 250 , \text{mL} = 0.250 , \text{L} )

Then, apply the formula:

( \text{grams of NaOH} = 0.250 , \text{L} \times 0.205 , \text{mol/L} \times 40.00 , \text{g/mol} )

( \text{grams of NaOH} = 2.05 , \text{grams} )

Therefore, 2.05 grams of NaOH are needed to prepare 250 mL of 0.205 M NaOH solution.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.