What mass of #"magnesium hydroxide"# is required to give a #1L# volume of #[Mg(OH)_2]# whose concentration is #0.01mol*L^-1# with respect to the hydroxide?

Answer 1

For a #1*L# volume of #0.01*mol*L^-1# #Mg(OH)_2#, #0.28*g# salt are required.

#"Molarity"# #=# #"Moles"/"Volume"# #=# #(0.01*mol)/(1.0*L)#.

And thus we need to dissolve #0.01*molxx58.32*g*mol^-1xx1/2# #=# #0.28*g# in a #1*L# volume. Why did I include the #1/2#?

However, #K_"sp",Mg(OH)_2=5.61×10^(−12)# at #298K#, which gives a solubility of #6.4xx10^-3*g*L^-1#. The question was thus not well-proposed.

Thus, magnesium hydroxide cannot provide such a concentration because it is too insoluble.

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Serenity Allen

To find the mass of magnesium hydroxide required, you can use the formula:

mass = molar mass * volume * concentration

The molar mass of magnesium hydroxide (Mg(OH)2) is 58.32 g/mol.

Substitute the values into the formula:

mass = 58.32 g/mol * 1 L * 0.01 mol/L = 0.5832 g

By signing up, you agree to our Terms of Service and Privacy Policy

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.

What mass of #"magnesium hydroxide"# is required to give a #1*L# volume of #[Mg(OH)_2]# whose concentration is #0.01*mol*L^-1# with respect to the hydroxide?

What mass of #"magnesium hydroxide"# is required to give a #1L# volume of #[Mg(OH)_2]# whose concentration is #0.01mol*L^-1# with respect to the hydroxide?