Molarity of a solution if 124.86 g of rbf are dissolved into a solution of water that has a final volume of 2.00L is 0.59.
<h3>What is molarity?</h3>
Molarity is used for dilute aqueous solutions held at a constant temperature. In general, the difference between molarity and molality for aqueous solutions near room temperature is very small and it won't really matter whether you use a molar or molal concentration.
MOLARITY = no of moles of solute/volume of soln in litres
No of moles of rbf = 124.6/104.46
= 1.19
Volume of soln = 2
Molarity=1.19/2 = 0.59
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Answer:
0.1472 mol/L is the concentration of the barium hydroxide solution.
Explanation:
Mass of potassium hydrogen phthalate = 2.050 g
Molar mass of potassium hydrogen phthalate = 
According to reaction , 2 moles of potassium hydrogen phthalate reacts with 1 mole of barium hydroxide, then 0.01004 moles of potassium hydrogen phthalate will :
of barium hydroxide
Moles of barium hydroxide = 0.005020 mol
Volume of the barium hydroxide solution = 34.10 mL = 0.03410 L
1 mL = 0.001 L
Molarity of the barium hydroxiude silution :
0.1472 mol/L is the concentration of the barium hydroxide solution.
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Three complete orders on each side of the m=0 order can be produced in addition to the m = 0 order.
The ruling separation is
d=1 / (470mm −1) = 2.1×10⁻³ mm
Diffraction lines occur at angles θ such that dsinθ=mλ, where λ is the wavelength and m is an integer.
Notice that for a given order, the line associated with a long wavelength is produced at a greater angle than the line associated with a shorter wavelength.
We take λ to be the longest wavelength in the visible spectrum (538nm) and find the greatest integer value of m such that θ is less than 90°.
That is, find the greatest integer value of m for which mλ<d.
since d / λ = 538×10⁻⁹m / 2.1×10 −6 m ≈ 3
that value is m=3.
There are three complete orders on each side of the m=0 order.
The second and third orders overlap.
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Answer:
"0.60 g" is the appropriate solution.
Explanation:
The given values are:
Volume of base,
= 30 ml
Molarity of base,
= 0.05 m
Molar mass of acid,
= 400 g/mol
As we know,
⇒ 
On substituting the values, we get
⇒ 
⇒ 
⇒ 
hence,
⇒ 
On substituting the values, we get
⇒ 
⇒ 
⇒ 
